$C'[0,1]$ is a Banach space with the norm $||f||=\max_{t\in[0,1]} \{|f(t)|,|f'(t)|\}$

Question

Let $C'[0,1]$ be the space of real functions defined in $[0,1]$ continuously differentiable in $(0,1)$ which derivative can be extended continuously to $[0,1]$. Show this is a Banach space with the norm $||f||=\max_{t\in[0,1]} \{|f(t)|,|f'(t)|\}$.

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My approach is really simple, since $$||f||=\max_{t\in[0,1]} \{|f(t)|,|f'(t)|\},$$ we can take $(f_n) \subseteq C'[0,1]$ a Cauchy sequence. Then for $N>0$ and $n,m>N$, $$||f_n-f_m||=\max_{t\in[0,1]} \{ |f_n(t)-f_m(t)|, |f_n'(t)-f_m'(t)| \}<\varepsilon.$$ In particular, $|f_n(t)-f_m(t)|<\varepsilon$, so $(f_n(t))$ is a Cauchy sequence for each $t\in[0,1]$. Then I basically repeat the argument used to prove that $C[0,1]$ is Banach space with the $\sup$ norm. Constructing $F(t)=\lim_{n\to\infty}f_n(t)$ for each $t\in[0,1]$ as the candidate function which $(f_n)$ converges.

Sadly, I'm struggling to conclude that $f$ is continuously differentiable in $(0,1)$. I suppose that I have to use the derivatives defined in the norm. However, a similar argument (considering $(f_n')$ a Cauchy sequence) as the one I used would yield to a $G$ function which $f_n'(t)\to G$. However, I cannot conclude $G=F'$.

Additionally, what does "which derivative can be extended continuously to [0,1]" exactly means? That I can define the limit of the derivative without breaking continuity?

Answer 1

$$\left|\frac{F(t+h)-F(t)}{h} - G(t) \right| =$$ $$ \left|\frac{F(t+h)-F(t)}{h} - \frac{f_n(t+h)-f_n(t)}{h} + \frac{f_n(t+h)-f_n(t)}{h} - f_n'(t) + f_n'(t) - G(t) \right| \leq $$ $$\left|\frac{F(t+h)-F(t)}{h} - \frac{f_n(t+h)-f_n(t)}{h}\right| + \left|\frac{f_n(t+h)-f_n(t)}{h} - f_n'(t)\right| + \left|f_n'(t) - G(t)\right|$$ $$\left|\frac{F(t+h)-F(t)}{h} - \frac{f_n(t+h)-f_n(t)}{h}\right| + \left|f_n'(\psi) - f_n'(t)\right| + \left|f_n'(t) - G(t)\right|$$ for some $\psi \in [t,t+h]$.

First choose $n$ such that for all $n \geq m$ we have, $$|f_n'(t)-G(t)| < \epsilon$$

Lets say $h$ in the above is chosen such that by uniform continuity of $f'_{m}(t)$ we have $$|f_m'(\psi) - f_m'(t)| < \epsilon, \forall \psi \in [t,t+h]$$.

Hence for all $n \geq m$, $$|f_n'(\psi) - f_n'(t)| \leq |f_n'(\psi) - f_m'(\psi) + f_m'(\psi) - f_m'(t) + f_m'(t)- f_n'(t)| \leq |f_n'(\psi) - f_m'(\psi)| + |f_m'(\psi) - f_m'(t)| + |f_m'(t)- f_n'(t)| < 5 \epsilon$$

Now with above chosen $h$, further choose $n$ such that for all $n \geq M_2 \geq m$, we have,

$$\left|\frac{F(t+h)-F(t)}{h} - \frac{f_n(t+h)-f_n(t)}{h}\right| < \epsilon$$

Hence we have, $$\left|\frac{F(t+h)-F(t)}{h} - G(t) \right| \leq $$ $$\left|\frac{F(t+h)-F(t)}{h} - \frac{f_n(t+h)-f_n(t)}{h}\right| + \left|f_n'(\psi) - f_n'(t)\right| + \left|f_n'(t) - G(t)\right| \leq 7 \epsilon$$

$m,M_2$ does not depend on $t$ by uniform continuity and uniform convergence.